What is the solution for y^{''}-y=sin(x)+e^x ?

The solution for y^{''}-y=sin(x)+e^x is y=c_{1}e^x+c_{2}e^{-x}+(-sin(x)+e^xx)/2

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y^{''}-y=sin(x)+e^x

y^{^{''}} - y = sin (x) + e^{x}

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{- sin ( x ) + e ^{x} x}{2}

Solution steps

y^{''} (x) - y = sin (x) + e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{- sin ( x ) + e ^{x} x}{2}

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{- sin ( x ) + e ^{x} x}{2}

y^{^{''}} + y = t^{2}

5 y^{^{''}} + 6 y^{^{'}} + 80 y = - 24 sin (4 t)

y^{^{''}} - y^{^{'}} - 2 y = e^{- x}

y^{^{''}} - 3 y^{^{'}} - 4 y = x

\frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} + 25 y = 30 x + 3

What is the solution for y^{''}-y=sin(x)+e^x ?
The solution for y^{''}-y=sin(x)+e^x is y=c_{1}e^x+c_{2}e^{-x}+(-sin(x)+e^xx)/2